Question

10)
△nop ≅△ by __
△nop ≅△...

Explanation:

Step1: Identify Triangle Parts

In the figure, \( \triangle NOP \) and \( \triangle NPQ \) (assuming the right - angled triangles share hypotenuse \( NP \), have one pair of equal sides (marked congruent) and right angles (\( \angle NOP=\angle NQP = 90^{\circ}\)).

Step2: Apply Congruence Criterion

For right - angled triangles, if the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the triangles are congruent (HL - Hypotenuse - Leg criterion). Also, we can see that \( NO = NQ \) (marked or from the figure's symmetry), \( NP \) is common (hypotenuse), and \( \angle NOP=\angle NQP = 90^{\circ}\). So \( \triangle NOP\cong\triangle NQP \) by HL (or we can also check for SAS: \( NO = NQ \), \( \angle NOP=\angle NQP \), \( OP = QP \) (marked congruent), so SAS is also possible. But looking at the right angles, HL is more direct for right triangles). Wait, from the figure, the two right - angled triangles are \( \triangle NOP \) and \( \triangle NPQ \) (with right angles at \( O \) and \( Q \)). The sides \( NO \) and \( NQ \) are equal (maybe from the figure's marking), \( OP \) and \( QP \) are equal (marked), and \( NP \) is common. So by SAS (since \( NO = NQ \), \( \angle NOP=\angle NQP = 90^{\circ}\), \( OP = QP \)) or by HL (since \( NP \) is hypotenuse, \( NO = NQ \) (leg) and \( OP = QP \) (leg)). But the more appropriate here is \( \triangle NOP\cong\triangle NQP \) by HL (or SAS). But looking at the standard, if we consider the two right triangles with hypotenuse \( NP \) and leg \( NO = NQ \) (or \( OP = QP \)), then the congruence is \( \triangle NOP\cong\triangle NPQ \) (wait, maybe the other triangle is \( \triangle NPQ \) with \( Q \) as the right - angled vertex). So the first blank is \( \triangle NPQ \) (or \( \triangle NQP \)) and the second blank is HL (or SAS). But from the figure's marking, the equal sides are the legs and the hypotenuse is common. So the congruence criterion is HL (Hypotenuse - Leg) or SAS (Side - Angle - Side).

Answer:

\( \triangle NOP\cong\triangle \boldsymbol{NPQ} \) by \(\boldsymbol{HL}\) (or \( \triangle NOP\cong\triangle \boldsymbol{NQP} \) by \(\boldsymbol{SAS}\))